Optimal. Leaf size=159 \[ \frac{\tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^n \left (\frac{b \tan (c+d x)}{a}+1\right )^{-n} F_1\left (\frac{3}{2};1,-n;\frac{5}{2};-i \tan (c+d x),-\frac{b \tan (c+d x)}{a}\right )}{3 d}+\frac{\tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^n \left (\frac{b \tan (c+d x)}{a}+1\right )^{-n} F_1\left (\frac{3}{2};1,-n;\frac{5}{2};i \tan (c+d x),-\frac{b \tan (c+d x)}{a}\right )}{3 d} \]
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Rubi [A] time = 0.221012, antiderivative size = 159, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3575, 912, 130, 511, 510} \[ \frac{\tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^n \left (\frac{b \tan (c+d x)}{a}+1\right )^{-n} F_1\left (\frac{3}{2};1,-n;\frac{5}{2};-i \tan (c+d x),-\frac{b \tan (c+d x)}{a}\right )}{3 d}+\frac{\tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^n \left (\frac{b \tan (c+d x)}{a}+1\right )^{-n} F_1\left (\frac{3}{2};1,-n;\frac{5}{2};i \tan (c+d x),-\frac{b \tan (c+d x)}{a}\right )}{3 d} \]
Antiderivative was successfully verified.
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Rule 3575
Rule 912
Rule 130
Rule 511
Rule 510
Rubi steps
\begin{align*} \int \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^n \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sqrt{x} (a+b x)^n}{1+x^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{i \sqrt{x} (a+b x)^n}{2 (i-x)}+\frac{i \sqrt{x} (a+b x)^n}{2 (i+x)}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{i \operatorname{Subst}\left (\int \frac{\sqrt{x} (a+b x)^n}{i-x} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{i \operatorname{Subst}\left (\int \frac{\sqrt{x} (a+b x)^n}{i+x} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac{i \operatorname{Subst}\left (\int \frac{x^2 \left (a+b x^2\right )^n}{i-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}+\frac{i \operatorname{Subst}\left (\int \frac{x^2 \left (a+b x^2\right )^n}{i+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=\frac{\left (i (a+b \tan (c+d x))^n \left (1+\frac{b \tan (c+d x)}{a}\right )^{-n}\right ) \operatorname{Subst}\left (\int \frac{x^2 \left (1+\frac{b x^2}{a}\right )^n}{i-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}+\frac{\left (i (a+b \tan (c+d x))^n \left (1+\frac{b \tan (c+d x)}{a}\right )^{-n}\right ) \operatorname{Subst}\left (\int \frac{x^2 \left (1+\frac{b x^2}{a}\right )^n}{i+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=\frac{F_1\left (\frac{3}{2};1,-n;\frac{5}{2};-i \tan (c+d x),-\frac{b \tan (c+d x)}{a}\right ) \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^n \left (1+\frac{b \tan (c+d x)}{a}\right )^{-n}}{3 d}+\frac{F_1\left (\frac{3}{2};1,-n;\frac{5}{2};i \tan (c+d x),-\frac{b \tan (c+d x)}{a}\right ) \tan ^{\frac{3}{2}}(c+d x) (a+b \tan (c+d x))^n \left (1+\frac{b \tan (c+d x)}{a}\right )^{-n}}{3 d}\\ \end{align*}
Mathematica [F] time = 1.03526, size = 0, normalized size = 0. \[ \int \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^n \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.243, size = 0, normalized size = 0. \begin{align*} \int \sqrt{\tan \left ( dx+c \right ) } \left ( a+b\tan \left ( dx+c \right ) \right ) ^{n}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (d x + c\right ) + a\right )}^{n} \sqrt{\tan \left (d x + c\right )}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \tan \left (d x + c\right ) + a\right )}^{n} \sqrt{\tan \left (d x + c\right )}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan{\left (c + d x \right )}\right )^{n} \sqrt{\tan{\left (c + d x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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